Finding the divided difference of $f(x)=x^{n+1}$ when I have only $n$ points $x_1,...,x_n$?

Question

The problem is to find $x^{n+2}[x_0,...,x_n]$ I managed to represent it as $(x_0+...+x_n) +x^{n+1}[x_1,...,x_n]$ using that $x^{n+2} = x^1x^{n+1}$ I have to find the divided difference $x^{n+1}[x_1,...,x_n]$ is this possible and if it is how do I do it?

Answer 1

I think you have a typo... when you apply "Leibniz" rule to $(x x^{n+1})[x_0, \cdots, x_n]$ you should get

\begin{align} (x x^{n+1}) [x_0,\cdots,x_n]=&x[x_0] x^{n+1}[x_0,\cdots,x_n] + x[x_0,x_1]\cdot x^{n+1}[x_1,\cdots,x_n]\\ =& x_0(x_0+ \cdots + x_n) + x^{n+1}[x_1,\cdots,x_n] \\ \end{align}

you can now perform the same reasoning for $x^{n+1}[x_1,\cdots,x_n]$, yielding \begin{align} (x x^{n}) [x_1,\cdots,x_n]=&x[x_1] x^{n}[x_1,\cdots,x_n] + x[x_1,x_2]\cdot x^{n}[x_2,\cdots,x_n]\\ =& x_1(x_1+ \cdots + x_n) + x^{n}[x_2,\cdots,x_n] \\ \end{align}

and so on... until you get the final answer: $$ x_0(x_0+ \cdots + x_n) + x_1(x_1+\cdots+x_n)+ \cdots + x_n^2 $$ or $$ \sum_{i=0}^n x_i^2+\sum_{i=0}^n \sum_{j=i+1}^n x_i x_j $$